Regarding the definition of the electrical potential energy

Question

Let us say we have two charge positive $q_{1}$ and $q_{2}$ very far away from each other. Suppose also that $q_{1}$ is stuck in its place. If we want to bring them closer to some distance $b$ apart, we would have to push $q_{2}$ with a certain force $\,\textbf{f}_{ext}\,$ against the repulsive coulombian force: $$\mathbf{f_{12}}=\frac{q_{1}q_{2}}{r_{12}^{2}}\mathbf{\hat{r}_{12}}$$

where $\mathbf{\hat{r}_{12}}$ is the unit vector pointing from $q_{1}$ toward $q_{2}$.

The potential energy of the final configuration ($b$ apart from each other) is equal to:$$ U=\int _{distance} -\textbf{f}_{12}\,.\textbf{ds}$$

In which we're integrating from the intial position of $q_{2}$ to its final position. This potential energy is by definiton equal to the work needed to displace $q_{2}$.

How could $q_{2}$ even be displaced if we are pushing it with a force $-\textbf{f}_{12}\,$ exactly equal in magnitude to the repulsive force $\textbf{f}_{12}\,$?

Or is that the potential energy $\,U$ is defined as:$$ U=\int _{distance} \textbf{f}_{ext}\,.\textbf{ds}$$

Where $\,\mathbf{f}_{ext}$ is higher in maghitude than $\,\textbf{f}_{12}\,$ by an infinitesimal amount?

Answer 1

You want to move the test charge quasistatically. That means, move it so slowly that its kinetic energy, and its interaction with magnetic fields don't affect your result.

So you only apply a force infinitesimally greater than ${\bf f}_{21}$, to move the charge at a very low speed.

Mathematically, you could consider applying a force $-(1+\epsilon){\bf f}_{12}$ and then obtain the potential energy as the limit of your integral as $\epsilon$ approaches 0.

Answer 2

If the force you were applying on $q_2$ were greater in magnitude than $\mathbf{f}_{12}$, then $q_2$ would accelerate, gaining kinetic energy. Part of the work done by the force you apply on $q_2$ would contribute to this kinetic energy. Then, the integral of this force along the trajectory of $q_2$ would not equal the potential energy, it would be greater because of this kinetic energy contribution. In addition, if there were other forces being applied on $q_2$ as it is being moved, then the integral of $\mathbf{f}_{ext}$ would no longer correspond to the electrostatic potential difference.

In truth, if there are only electrostatic forces being applied on $q_2$ (besides the external force), you can integrate $\mathbf{f}_{ext}$ along the trajectory of $q_2$ to find the potential energy difference. It doesn't matter what $\mathbf{f}_{ext}$ is, as long as the initial and final speeds (and thus kinetic energies) of $q_2$ are equal (usually zero). If $q_2$ were approaching $q_1$ radially with a certain initial speed, setting $\mathbf{f}_{ext} = -\mathbf{f}_{12}$ would keep $q_2$ moving with the same velocity until its final position. Thus, integrating $\mathbf{f}_{ext}$ will give the same result as integrating $-\mathbf{f}_{12}$.